Calculus:Nth Differential Coefficient of Standard Functions

8/17/2019 Calculus:Nth Differential Coefficient of Standard Functions

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SUCCESSIVE DIFFERENTIATION:

LEIBNITZ'S THEOREM

OBJECTIVES

At the end of this session, you will be able to understand:

Definition

nth

Differential Coefficient of Standard Functions

Leibnitz’s Theorem

DIFFERENTIATION: If be a differentiable function of x, then)( x f y = )(' x f dx

dy= is called the first

differential coefficient of y w.r.t x.

Hence, differentiating both side w.r.t. x, we have

( ) ( )

( )

( )

2 2

2 2

2 3 3

2 3 3

' '' .

d dy d d be represented by ; then ''

dx dx

d dSimilarly is represented by ; ie ''' and so on

dx dx

T

d dy d f x f x

dx dx dx

y y Let f x

dx dx

y d y d y f x

dx dx

= =

=

=

2 3

2 3

n

1 2 3

he expressions , , , ....... are called the first, second, third, .....nth differen-

tial coefficient of y.

These function are usually written as

y',y'',y''',......y or y ,y y ......

n

n

dy d y d y d y

dx dx d x dx

2 3

n..................y and also , , ,....... .n

Dy D y D y D y

nth

DIFFERENTIAL COEFFICIENT OF STANDARD FUNCTION:

(i) Differential Coefficient of :m

x

m 1

1

2 3

2 3

n

y = x , then y ;

y ( 1) ; y ( 1)( 2) and so on

In general, y ( 1)( 2)( 3).........................( 1) ;

m

m m

m n

If mx

m m x m m m x

m m m m m n x

−

− −

−

=

= − = − −

= − − − − +

1


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Note: If m be a positive integer, we have

ny 1.2.3............ !;m m= =

Hence ( ) ( 1)( 2)( 3)............( 1)m mn D x m m m m m n x n−= − − − − +

(ii) Differential Coefficient of ( )max b+ :

m 1

1

2 2 3 3

2 3

If y = (ax+b) , then ( ) ;

( 1)( ) ; ( 1)( 2)( ) and so on.

In general ( 1)( 2)( 3)............( 1)( )

m

m m

n m n

n

y am ax b

y a m m ax b x y a m m m ax b x

y a m m m m m n ax b x

−

− −

−

= +

= − + = − − +

= − − − − + +

[Note: In the first differentiation, the last term in it is ( )11+−m

)2

; in the second differentiation it is

i.e. ( ;in the third differentiation it is ()1( −m )12 +−m −m i.e. )13( +−m .So the nth differentiation it

will be .])1+− n(m

Hence ( ) ( 1)( 2)( 3)............( 1)( )

!or ( ) ( )

( )!

In case m is negative integer,let , where is positive integer, then

( ) ( )( 1)( 2)....

m n m n

m n m n

p n

D ax b a m m m m m n ax b

m D ax b a ax b

m n

m p p

D ax b a p p p

−

−

−

+ = − − − − + +

+ = +−

−

+ = − − − − −

1

................[ ( 1)]( )

( 1) ( 1)( 2)....................[( 1)]( )

( 1)!( 1) ( )( 1)!

Note1. If .then ( ) !

Note2. If 1, we have ( ) (

p n

n p nn

n n p n

n p n

n

p n ax b

p p p p n ax ba

p n a ax b p

m n D ax b a

m D ax b

− −

− −

− −

−

−

− − − +

= − + + − + +

− += − +−

= + =

= − + = 1

1 1

1

1)( 1 1)................( 1 1) ( )

( 1) !( 1) 1.2.3........... ( ) ! ( ) .( 1)

( )

n n

n nn n n n nn

n

n a ax b

n ana ax b n a ax b

ax b

− −

− − − −

+

− − − − − + +

−− + = + =−

+

(iii) Differential Coefficient of ( )ax b+ :

1

1

2 2 3 3

2 32 2 3 3

4 43

4 4 4

(0!)If log( ), ( )( ) ( )

.1 .(1!) .2 .(2!) . .

( ) ( ) ( ) ( )

2.3 .(3!) and so on.( 1) .

( ) ( )

a a y ax b then y a ax bax b ax b

a a a a y y

ax b ax b ax b ax b

a a y

ax b ax b

−

= + = = + =+ +

= = − = = −+ + + +

= = −+ +

2


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1

1

1

.( 1)!In general, ( 1) .

( )

.( 1)!

Hence log( ) ( 1) . .( )

( 1) ( 1)! Note: log .

nn

n n

nn n

n

nn

n

a n y

ax b

a n

D ax b ax b

n D x

x

−

−

−

−= −

+

−

+ = − +

− −=

(iv) Differential Coefficient of :bxa

2 2

1 2

3 3

3

n

If , then log , (log ) .

, and so no.(log )

In general, ,(log )

Hence D . (log ) .

bx bx bx

bx

n nbxn

bx n nbxe

y a y ba a y b a a

y b a a

y b aa

a b aa

= = =

=

==

(v) Differential Coefficient of :axe

2 3 4

1 2 3 4

If , then

, , , and so on.

In general, , Hence .

ax

ax ax ax ax

n ax n ax n ax

n

y e

y ae y a e y a e y a e

y a e D e a e

=

= = = =

= =

(vi) Differential Coefficient of sin( ) :ax b+

1

2 22

1

If sin( ), then

cos( ) sin ( )2

2 cos( ) sin ( ) ,and so on.

2

1

In General, .sin 2

1Hence, ( ) sin

2

Note: sin

n

n n

n

y ax b

y a ax b a ax b

y a ax b ax ba

y ax b na

D ax b a ax b n

D x

π

π

π

π

= +

= + = + +

= − + = + +

= + +

+ = + +

sin2

n x

π = +

3


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(vii) Differential Coefficient of cos( ) :ax b+

1

2 22

If cos( ), then

sin( ) cos2

2sin cos ,and so on

2 2

1In general, cos .

2

1Hence, cos ( ) cos .

2

Note : cos co

n

n

n n

n

y ax b

y a ax b a ax b

y a ax b ax ba

y a ax b n

D x ax b a ax b n

D x

π

π π

π

π

= +

= − + = + +

= − + + = + +

= + +

+ = + +

=1

s .2

x nπ

+

(viii) Differential Coefficient of sin( ) and cos( ) :ax axe bx c e bx c+ +

1

2 2 2 1

1

If sin( ), then

sin( ) cos( ) [ sin( ) cos( )]

Putting cos and sin , we get

sin( ), where and tan ,

similary

ax

ax ax ax

ax

y e bx c

y e bx c be bx c e a bx c b bx c

a r b r

b y re bx c r a ba

φ φ

φ φ −

= +

= + + + = + +

= =

= + + = + =

+

21

1

2 2 12

1

2 2 12

, ,and so on.sin( 2 )

Hence, sin( ) sin( )

where ( ) and tan .

Similarly, cos( ) cos( )

where ( ) and tan .

ax

n ax n ax

n ax n ax

y r bx ce

D e bx c r bx c ne

br a b

a

D e bx c r bx c ne

br a ba

φ

φ

φ

φ

φ

−

−

= + +

+ = + +

= + =

+ = + +

= + =

Example. Find the derivative ofthn .cossin cxbxe ax

Solution.

4


5/14

{ }

2 2 / 2 1

/ 22 2

1 Let sin cos (2sin cos )

2

1 1 [sin( ) sin )] [ sin sin( ) ].( ( )

2 2

Now sin( ) ( ) sin tan

( ) si1

2

ax ax

ax ax ax

n n axax

nax

n

y e bx cx e bx cx

e bx cx x e x e b cbx c b c

b D bx cx b c e bx c nea

a b c e

y

−

= =

= + + = + −− +

+ = + + +

+ +

x

∴ ={ }

{ }

1

/ 22 2 1

n tan ( ) /( )

( )( ) sin tan( )

nax

x n b c ab c

b ca b c e x nb c

a

−

−

+ ++ −

+ − − +

Example. If .0)(2that proved ,sin22

12 =++−= ybaay ybxe y ax

Solution.

1

2 2

2

2

1

2 2

Let sin , sin cos

and sin 2 sin ...............(1)

Also 2 2 sin 2 cos ...............(2)

and ( )

ax ax ax

ax ax ax

ax ax

y e bx y ae bx be bx

y a e bx abe b e bx

ay a e bx abe bx

a b y a

= ∴ = +

= + −

− = − −

+ = 2 2

2 2

2 1

sin sin .................(3)

Adding (1),(2)and (3), we get

y - 2ay ( ) 0

ax axe bx b e bx

a b y

+

+ + =

LEIBNITZ'S THEOREM:

The find nth

differential coefficient of two function of x

If u and v are any two functions of x such that all their desired differential coefficients exist, then the nth

differential coefficient of their product is given by

1 1 2

1 2

1 2 2

( ) ( ). . . ....... . ...... .

or

( 1)( ) ( ). . .............

2!

n n n n n n n n r r

r

n n n n

n D uv D u v c D u Dv c D u D v c D n D v uD v

n n D uv D u v nD u Dv D uD v n

− − −

− −

= + + + + + +

−= + + + + 1 .n DuD v uDv− +

Proof.

5


6/14

Let , we have

( ) ( ) ( ). . ........ ......(1)

From (1) we see that the theorem is true for n 1.

Now assume that the theorem is true for a particular val

n

y uv

Dy uv D u Du v u Dv

=

= = +

=

1 2 2

1 2

1 1

1

ue of n,we have

( ) ( ). . . ....... .

...... ............(2)

Differentiating both siede of (2

n n n n n n r r

c c r

n r r n

r

D uv D u v n D u Dv n D u D v c D u D v

n D uD v uD v

− − −

− − ++

= + + + +

+ + +

( )

( ) (

( )

1 1 1 2

1 1

1 2 2 3 1 1

2 2

1 1 2

1 1

) w,r,t,x, we get

( ) ( ). . .

. . ..... . .

. . ..... .

n n n n n

c c

n n n r r n r

c c cr cr

n r r n r r n

cr cr

D uv D u v D uDv n D u Dv n D u D v

n D u D v n D u D v n D u D v n D u D v

n D u D v n D u D v DuD v u

+ + −

− − − + −

− + − − +

+ +

= + + +

+ + + + +

+ + + + +

( )

1

1 1 1 2

1 1 2

1 1

1

1

.

Rearranging the term, we get

( ) ( ). (1 ) ( ) ( ) .

..... ( )( . ) ...... ...(3)

But we know that

r

n n n n

c c c

n r r n

cr r

n n

D v

D uv D u v n D uDv n n D u D v

n n D u D v uD v

c c

+

+ + −

− + ++

= + + + + + +

+ + + +

+ 11 1

1 1

1 0 1 1 1 2 2,

1 1 1 1 1 2

1 2

1 1

1

. Therefore

1 , and so on.

Hence(3) gives

( ) ( ). ( ). ( ).( ) .........

..... . .....

n

r r

n n n n n n n

n n n n n n

n n r r

r

c

c c c c c c c

D uv D u v c D u Dv c D u D v

c D u D v

++ +

+ +

+ + + + −

+ − ++

=

+ = + = + =

= + + +

+ + + 1. . ..............(4)nu D v+

)r +

From (4) we see that if the theorem is true for any value of n, it is also true for the next value of n. But

we have already seen that the theorem is true for n =1.Hence is must be true for n =2 and so for n =3,

and so on. Thus the Leibnitz's theorem is true for all positive integral values of n.

Example. Find the nth differential coefficients of

(i) sin cos ,

(ii) log[( )( )].

ax bx

ax b cx d + +Solution.

6


7/14

1 1 (i) Let sin cos [2sin cos ] [2sin( ) sin( ) ].

2 2

1we know that sin( ) sin .

2

1 1( ) sin ( ) ( ) sin ( ) .

2 2

n n

n n n

y ax bx ax bx a b x a b

D ax b a ax b n

y a b a b x n a b a b x n

π

π π

= = = + + −

+ = + +

1

2

x

∴ = + + + + − − +

[ ]n 1

1 1

1

(ii) Let log ( )( ) log( ) log( ).

We know that log( ) ( 1) ( 1)! ( )

( 1) ( 1)! ( ) ( 1) ( 1)! ( )

( 1) ( 1)! .

( ) ( )

n n n

n n n n n

n

n nn

n n

y ax b cx d ax b cx d

D ax b n a ax b

y n a ax b n c cx

a cn

ax b cx d

− −

− − −

−

= + + = + + +

+ = − − +nd −∴ = − − + + − − +

= − − +

+ +

Example. Find the nth derivatives of 2

1.

1 5 6 x x− +

Solution.

2

2

1 1

1

1 1Let .

(2 1)(3 1)6 5 1

1 (3 1) (2,

2 1 3 1 (2 1)(3 1)6 5 11 1

Putting ,1 , i.e. 3 ; putting , 2.2 3 2

2 3Hence 2(2 1) 3(3 1)

2 1 3 1

Therefor 2(2 1)n

n n

y x x x x

A B A x B x

x x x x x x B

x B x A

y x x x x

d y x

dx

− −

−

= =− −− +

− + −1)∴ ≡ + ≡

− − − −− +

= = − = − = =

= + = − − −− −

= − 1

1 1

1 1

1 1

1 1

3(3 1)

Now we apply the formula,

( ) ( 1) ( !)( ) .

Hence 2.2 ( 1) ( !)(2 1) 3.3 ( 1) ( !)(3 1) .

2 3or ( 1) ( !) .

(2 1) (3 1)

n

x

n

n n n n

n n n n n n

n nn

n n n

d x

dx

D ax b n ax b a

y n x n x

y n x x

−

− − −

− − − −

+ +

+ +

− −

+ = − +

= − − − − −

= − +

− −

7


8/14

Example.1 /2

If sin cos , prove that 1 ( 1) sin 2 .n n n y ax ax y a ax = + = + −

1 /22

1/ 2

Let sin cos , then

1 1

sin cos2 2

1 1 sin cos

2 2

1 1 1 2sin cos

2 2

[1 sin(2

n n

n

n

n

n

y ax ax

y a ax n a ax n

a ax n ax n

a ax n ax n

a a

π π

π π

π π

= +

∴ = + + +

= + + +

= + + +

= + 1 /2 1/ 2

1/ 2

)] [1 sin cos 2 cos sin 2 ]

[1 ( 1) sin 2 ] [ sin 0 and cos ( 1) ]n n nn

n n ax n

y a ax n n

π π π

π π

+ = + +

= + − = = −Q

ax

Example.

2 22 2 2 2 2

2 3If cos sin , prove that .

d p a b p a b p

d pθ θ

θ

= + + =

2

Solution.

2 2 2 2 2

2 2

Given cos sin ...(1)

Differentiating both sides of (1) w.r.t , we get

2 2( )cos sin ...(2)

Aga

p a b

dp p b a

d

θ θ

θ

θ θ

θ

= +

∴ = −

222 2 2 2

2

2

in differentiating both sides of (2) w.r.t , we get

( )(cos sin ) ...(3)

Multiplying (3) by and substituting the value of form (1) a

d p dp p b a

d d

dp p

d

θ

θ θ θ θ

θ

+ = − −

23 2 2 2 2 2 2 2 2 2 2

2

23 2 2 2 2 2 2 2 2 2 2 2 2 2

2

nd (3), we get

( ) cos sin ( )(cos sin )

or ( cos sin )( )(cos sin ) ( ) cos sin

d p p b a p b a

d

d p p a b b a b ad

θ θ θ θ θ

θ θ θ θ θ θ θ θ

+ − = − −

= + − − − −

8


9/14

24 3 2 2 2 2 2 2 2 2 2 2 2 2

2

2 2 2 2 2

or ( )[(cos sin )( cos sin ) ( )cos sin ]

( cos sin )

d p p p b a a b b a

d

a b

θ θ θ θ θ θ

θ θ

+ = − − + − −

+ −

θ

2 2 2 4 2 4 2 2 2 2 2

2 2 2 2 2 2

2 2 2

2 3

=( )[( cos sin ) ( cos sin )

= (cos sin )

Hence .

b a a b a b

b a a b

d p a b p

d p

θ θ θ θ

θ θ

θ

− − + +

+ =

+ =

Example.1

Find if log .n

n y y x −= x

Solution.

1 1 1 1 2 1

3 1 3 1

By Leibnitz's theoram, we get

( 1)( log ) ( ) log ( ) log ( ) log2!

( 1)( 2) ( ) log ....... .

3!

! Now ;

( )!

n n n n n n n nn

n n n n

n m m n n n

n n y D x x D x x nD x D x D x D

n n n D x D x x D Logx

m D x x D x

m n

− − − − − −

− − −

−

−= = + +

− −+ + +

=−

2 x

1

1 1 1 1

2 1 3 1 2

1

2

0

! ; ( 1)!

( 1)!

( 1)! ( 1)! ;

1! 2!

( 1)!and log ( 1)

Hence

1 ( 1) ( 1)! ! ( 1)( 2) ( 1)!( 1)!

2! 1! 3! 2!

n m m n n n

n n n n

n n

n

n

m D x x D x n

m n

n n D x x D x x

n D x

x

n n n n n n nn n x

x x y

−

− − + − −

− − − −

−

=

= ∴ = −− +

− −= =

−= −

− − − − − − + − +

=

2

3

11

1

1 2 3

2....

( 1) ( 1)! .......

( 1)! = [1 {1 ...... ( 1) }]

( 1)! =

nn

n

n n n n

n

x x

n x

x

nc c c c

x

n

−−

+

+

− −

+

−− − − − + + −

− ( 1)![1 (1 1)n

n −− − =

9


10/14

1 2

1

1 1 1

1

2 1 1 1

1

1 1

Aliter. log ( 1) log .

( 1) log ( 1) .

Differentiating both sides ( 1)times, we have

( ) ( 1) .

( 1) ( 1) (

n n

n n n

n n n n

n n n

y x x y n x x x

xy n x x x n y x

n

D xy n D y D x

xy n y n y

− −

− − −

− − − −

− +

= ∴ = − +

∴ = − + = − +

−

= − +

2n−

∴ + − = − +( 1)

1)! or nn

n y!

x

−− =

Example.2

2 1 y cos(log ) sin(log ), show that 0 If a x b x x y xy y= + + +2 2

2 1and (2 1) ( 1) 0.n n n x y n xy n y+ ++ − + + =

=

Solution.

1 1

2 1

2

2 1

Let cos(log ) sin(log ),1 1

sin(log ). cos(log ) or sin(log ) cos(log )

Now again differentiating both sides, we get

1 1 cos(log ). sin(log )

or [ cos(log )

y a x b x

y a x b x xy a x b x x x

xy y a x b x x x

x y xy a x b

= +

= − + = − +

+ = − −

+ = − +2

2 1

2

2 1

2

2 1

2 2 1 2 2 2 1

2 2 2 1 1

sin(log )]

or

or 0.

Again differentiating both sides in times by Leibnitz's theorem,

( ) ( ) ( ) 0.

( 1)or + 0

2

or

n n n

n n n n n

n

x

x y xy y

x y xy y

D x y D xy D y

n n x D y nDx D y D x D y xD y nD y y− − +

+ = −

+ + =

+ + =

−+ + +

2

2 1 1

2 2

2 1

2 ( 1) 0

or (2 1) ( 1) 0.

n n n n n n

n n n

x y nxy n n y xy ny y

x y n xy n y

+ + +

+ +

+ + − + + + =

+ − + + =

+ =

x 2

Example If prove that1sin( sin ). y m −= 2 2 1(1 ) 0 x y xy m y− − + =2

and deduce that

2 2

2 12 1) (n nn xy n+ ++ −(1 ) ( ) 0. x y m y− − − n =

x

Solution: Let 1sin( sin ). y m −=

10


11/14

1 2 2 2

1 12

2 2 2 2 2 1 2 2 2

1

2 2 2 2

1

2 2 2

1 2 1 1 2

cos( sin ). . or (1 ) cos ( sin ).(1 )

(1 ) sin ( sin )

(1 ) .Again differentiating both sides, we have

2 (1 ) 2 2 0. or

m y m x x y m m

x

or x y m m m x m m y

x y m y m

y y x xy m yy y

− −

−

= − =−

− = − = −

∴ − + =

− − + =

2 1 x

2 2

1

2 2

2 1 1

2 2 2

2 1

(1 ) 0.

Now differntiating n time by Leibnitz's theorem, we get

( 1)(1 ) ( 2 ) ( 2) 0,

2!

or (1 ) (2 1) ( ) 0.

n n n n n n

n n n

x xy m y

n n y x ny x y xy ny m y

x y n xy n m y

+ + +

+ +

− + =

−− + − + − − − + =

− − + − − =

To find The nth Derivative When x = 0

Example: Find ( 10) . if sin( sin ).n y y a −= x

x

Solution:

Let ……..(1)1sin( sin ). y a −=

1

12

2 2 2 2 1 2 2 2 1 2 2

1

2 2 2 2 21

cos( sin ). ,(1 )

or (1 ) cos ( sin ) sin ( sin )

or (1 ) 0. ................(2)

Differentiating (2), w

a y a x

x

y x a a x a a a x a a y

y x a y a

−

− − 2

∴ =−

− = = − = −

− + − =

2 2 2

1 2 1 1

2 2

2 1 1

2

2 1 1

e have

2 (1 ) ( 2 ) 2 0.

(1 ) 0 ...................(3)

Differentiating (3) n times, we have

( 1)(1 ) 2 .2

2!n n n n

y y x y x a yy

or y x xy a y

n n y x ny x y xy n+ + +

− − − + =

− + + =

−− − − − − 2

2 2 2

2 1

0.

or (1 ) (2 1) ( ) 0. ..................(4)

n n

n n n

y a y

y x n xy n a y+ +

+ =

− − + − − =

Putting x = 0 in (1), we get (y)0 = 0.

Putting x = 0 in (2), we get (y1)0 = 0

Putting x = 0 in (3), we get (y2)0 = 0 and

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Putting x = 0 in (4), we get2 2

2 0 0( ) ( )(n n y n a+ = − ) y

.

.

.

a

.

0n

Now putting n = 2 in (5),2 2

6 0 2 0( ) (2 )( ) 0 y a y= − =

Putting n = 4 in (5), 2 26 0 4 0( ) (4 )( ) 0 y a y= − =

Similarly 8 0( ) 0 y = Thus the derivatives for which n is even are zero

Again, putting2 2 2 2

3 0 1 01,( ) (1 ).( ) (1 ) .n y a y a= = − −

Now when n is odd.2 2

2 0 0( ) ( )( )n n y n a y+ = −

Putting n is place of (n-2) we obtain

[ ]

2 2

0 2 0

2 2 2 2 2 2 2 2

3 0

2 2 2 2 2 2 2 2

( ) ( 2) ( )

( 2) ( 4) ( 6) ......... 3

( 2) ( 4) ......... 3 1 . .

n n y n a y

n a n a n a a y

n a n a a a a

− = − −

= − − − − − − −

= − − − − − −

Example.1 2

2 1

2

2 1

If tan , prove that (1 ) 2 0 and deduce that

(1 ) 2( 1) ( 1) 0 Hence determine ( )n n n

y x x y xy

x y n xy n n y y

−

+ +

= + + =

+ + + + + =

Solution.1

1 2

2

1

Let tan .........(1)

1 , ...(2)

(1 )

or (1 ) 1 0. ...(3)

Differentiating (3), we

y x

y x

x y

−=

∴ =+

+ − =2

2 1

2

2 1 1

2

2 1

get (1 ) 2 0 .........(4)

Now , differentiating (4) n times by Leibnitz's theorem, we get

( 1) (1 ) (2 ) .2 2 2 0

2!

or (1 ) 2( 1) ( 1)

n n n n n

n n

x y xy

n n y x ny x y xy ny

x y n xy n n y

+ + +

+ +

+ + =

−+ + + + + =

+ + + + + 0 .........(5)n =

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[ ]

0 1 0 2 0

2 0 0

Putting x = 0, in (1),(2) and (4), we get

( ) 0, ( ) 1, ( ) 0.

Also putting x = 0 in (5) we get

( ) ( 1) ( ) . .............(6)

Putting n-2 in

n n

y y y

y n n y+

= = =

= − +

0 2 0

4 0

n-2 0 4 0

place of n in the foumula (6),we get

( ) [( 1)( 2)]( )

[ {( 1)( 2)}][ {( 3)( 4)}]( )

Since from (6), we have (y ) {( 3)( 4)}]( )

Case I. When n is even, we h

n n

n

n

y n n y

n n n n y

n n y

−

−

−

= − −

= − − − − − −

= − − −

0 2 0

2 0

0

ave

( ) [ {[( 1)( 2)}][ {( 3)( 4)}]....[ (3)(2)]( )

= 0, Since ( ) 0.

Case II. When n is odd, we have ( ) [ {[( 1)( 2)][ {( 3)( 4)}]....[ (4)(3)]

n

n

y n n n n y

y

y n n n n

= − − − − − − −

=

= − − − − − − − 1 0( 1)2

1 0

[ (2)(1)( )

( 1) ( 1)!, since (y ) 1.n

y

n−

−

= − − =

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ADDITIONAL PROBLEMS:

1. Find the nth differential coefficient of

( ) ( )

3

2

2

( ) sin

( ) sin cos3

( ) cos sin

x( )

2 2 3

ax

i x

ii x x

iii e x x

iv x x+ +

2. If ( )2 22 1sin , prove that 2 0ax bx y ay a b y= − + y e + =

3. If ( ) ( )1 2 2 1cos sin , prove that 1 0 x x y xy m y−= − − 2+ = y m and

( )2 2

2 11 (2 1) ( )n n x y n xy m n y+ +− − + + −

2

0n =

4. If ( ) ( )2

1 2

2 1sin , prove that 1 2 0 y x x y xy−= − − − = and deduce that

( )2 22 11 (2 1)n n x y n xy n y+ +− − + − 0n =

5. If (1tan

, prove that x

y e−

= )2 2 11 {2( 1) 1} ( 1)n n x y n x y n n y+ ++ + + − + + 0n =

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Calculus:Nth Differential Coefficient of Standard Functions